Ali Barber, the carpet merchant, has a rectangular piece of carpet whose dimensions are unknown. Unfortunately, his tape measure is broken and he has no other measuring instruments. However, he finds that if he lays it flat on the floor of either of his storerooms, then each corner of the carpet touches a different wall of that room. He knows that the sides of the carpet are integral numbers of feet and that his two storerooms have the same (unknown) length, but widths of 38 feet and 50 feet respectively. What are the carpet dimensions?
Problem
Source: IMO Shortlist 1989, Problem 3, ILL 4
Tags: calculus, algebra, rectangle, equation, area, IMO Shortlist
Amir Hossein
15.03.2010 21:32
Sorry,anyone can help me solving this problem? tanks.
pco
17.03.2010 21:37
amparvardi wrote: Sorry,anyone can help me solving this problem? tanks. 1) What are the rules for including a $ (x,y)$ carpet in a $ (L,a)$ room ... ? ============================================== ... such that each corner of the carpet is on a different wall. Let $ \theta$ the angle between carpet side $ x$ and wall $ L$. We get : $ L=x\cos(\theta)+y\sin(\theta)$ $ a=x\sin(\theta)+y\cos(\theta)$ So $ \cos(\theta)=\frac{xL-ay}{x^2-y^2}$ and $ \sin(\theta)=\frac{yL-ax}{y^2-x^2}$ And so the condition is $ (xL-ay)^2+(yL-ax)^2=(x^2-y^2)^2$ Which may be written $ \boxed{(x^2+y^2)(L^2+a^2)-4aLxy=(x^2-y^2)^2}$ 3) resolution of the problem ================== Let $ (x,y)$ be the dimension of the carpet (integers) Let $ (L,a=50)$ and $ (L,b=38)$ the dimension of the two rooms. We get : $ (x^2+y^2)(L^2+a^2)-4aLxy=(x^2-y^2)^2$ $ (x^2+y^2)(L^2+b^2)-4bLxy=(x^2-y^2)^2$ Subtracting, we get $ (x^2+y^2)(a+b)=4Lxy$ and so $ L=\frac{(x^2+y^2)(a+b)}{4xy}$ Plugging this in the first equation,we get $ (x^2+y^2)(\frac{(x^2+y^2)^2(a+b)^2}{16x^2y^2}+a^2)-(x^2+y^2)(a^2+ab)=(x^2-y^2)^2$ So $ (x^2+y^2)((x^2+y^2)^2(a+b)^2-16abx^2y^2)=16x^2y^2(x^2-y^2)^2$ So $ (x^2+y^2)((x^2+y^2)^2(a+b)^2-16abx^2y^2)=16x^2y^2((x^2+y^2)^2-4x^2y^2)$ Let $ x=pu$ and $ y=pv$ with $ \gcd(x,y)=p$ and so $ \gcd(u,v)=1$. The equation becomes : $ (u^2+v^2)((u^2+v^2)^2(a+b)^2-16abu^2v^2)=16p^2u^2v^2((u^2+v^2)^2-4u^2v^2)$ So $ u^2v^2 | (u^2+v^2)^3(a+b)^2$ and so $ uv|a+b$ (remember $ \gcd(u,v)=1$) Wlog say $ u> v$, we get $ uv|88$ and so $ (u,v)\in\{(2,1),(4,1),(8,1),$ $ (11,1),(22,1),(44,1),$ $ (88,1),(11,2),(11,4),(11,8)\}$ The equation $ (u^2+v^2)((u^2+v^2)^2(a+b)^2-16abu^2v^2)=16p^2u^2v^2((u^2+v^2)^2-4u^2v^2)$ allows to find $ p^2$ from $ (u,v)$ : Applying this to the $ 10$ couples we got, only one gives a perfect square : $ (2,1)$ gives $ p^2=625$ Hence the unique solution $ \boxed{(x,y)=(50,25)}$ which implies $ L=55$ (and so is an integer too, which was not a condition of the problem) and it is easy to check back that these numbers match the requirements.